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The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Frady. The amount of ${PbSO}_{4}$ electrolysed in grams during this process is: (Molar mass of ${PbSO}_{4} = 303 g {mol}^{- 1})$

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22.8

11.2

7.6

11.4

answer is C.

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Detailed Solution

{PbSO}_{4} (s) + {2 e}^{-} \to Pb (s) + {SO}_{4}^{2 -} For 2 F current passed {PbSO}_{4} electrolysed = 303 gms {mol}^{- 1} For 0.05 F, {PbSO}_{4} electrolysed = \frac{0.05 \times 303}{2} = 7.6 gm

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